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[ Python exercise: last letter / first letter ]

I'm a newbie Python student and I'm going through some simple (but now, for me, complicated) exercises. I tried in many ways, but I decided to stop guessing, because I believe it won't be a sane routine of learning.

I have to solve the following exercise:

Write a lastfirst(lst) function, that, given a list, returns the first word in the list that starts with a different character from the last character of the previous word. If there isn't such a word, return None.


lst = ['sole','elmo','orco','alba','asta'] returns 'alba'

lst = ['sky','you','use','ear','right'] returns None

I've tried to solve it, and what I had is this:

lst = ['sole','elmo','orco','alba','asta']

def lastfirst(lst):

    cont = 0
    d = 1 
    for a in lst[cont:]: 
        for b in lst[d:]:       
            if a[-1] != b[0]:
                return lst[d]
                 cont = cont + 1   
                 d = d + 1  


The problem I detected is:

The program doesn't make a distinction between taking the first letter of the first word and the last letter of the second word, or the last letter of the first word and the first letter of the second word.

PS: Sorry for my English :)

Answer 1

I think it will work (in python 3):

lst = ['sole','elmo','orco','alba','asta']

def lastfirst(lst):
    for i in range(len(lst)-1):
        if lst[i][-1] != lst[i+1][0] :
            return lst[i+1]
    return None



Explanation(Modifications needed in your code) :

  • We don't need two for loops we can do it in single for loop.
  • Although you do increment cont variable in else statement but it will always compare it with same string which is a.

Another Input :

lst = ['sky','you','use','ear','right']



Hope it will help you.

Answer 2

You would use a double for loop when you need to test every word in lst against every other word in lst, but that's not what we want here. We just need a single for loop, and we need to store the previous word so we can test it against the current word. Like this:

def lastfirst(lst):
    if not lst:
        return None
    prev = lst[0]
    for word in lst[1:]:
        if word[0] != prev[-1]:
            return word
        prev = word
    return None

data = [
    ['sole', 'elmo', 'orco', 'alba', 'asta'], 
    ['sky', 'you', 'use', 'ear', 'right'],
    ['help', 'please', 'everybody', 'thanks'],

for lst in data:



My function first does

if not lst:
    return None

so we return immediately if we get passed an empty list. Otherwise, the program will crash when it attempts to do prev = lst[0]

Here's an efficient way to do the test using a single line.

def lastfirst(lst):
    return next((v for u, v in zip(lst, lst[1:]) if u[-1] != v[0]), None)

This code is obviously more compact than my previous version, and it might be a little faster. But it is harder to understand, especially if you are new to Python. Some people think that "one-liners" like this are more Pythonic, but actually it is more Pythonic to make your code as readable as possible. :)

Answer 3

Here's a solution using itertools.

First, define a function that returns a boolean if your condition is met:

def check_letters(apair):
"In a pair, check last letter of first entry with first letter of second"
    return apair[0][-1] == apair[1][0]

Now we use the pairwise function from the itertools module recipes:

import itertools
def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = itertools.tee(iterable)
    next(b, None)
    return itertools.izip(a, b)

And finally:

lst = ['sole','elmo','orco','alba','asta']
lstlast = [item[1] for item in pairwise(lst) if not check_letters(item)]
# returns ['alba']